with - sizeof() c++




Why does “sizeof(a ? true:false)” give an output of four bytes? (5)

I have a small piece of code about the sizeof operator with the ternary operator:

#include <stdio.h>
#include <stdbool.h>

int main()
{
    bool a = true;
    printf("%zu\n", sizeof(bool));  // Ok
    printf("%zu\n", sizeof(a));     // Ok
    printf("%zu\n", sizeof(a ? true : false)); // Why 4?
    return 0;
}

Output ( GCC ):

1
1
4 // Why 4?

But here,

printf("%zu\n", sizeof(a ? true : false)); // Why 4?

the ternary operator returns boolean type and sizeof bool type is 1 byte in C.

Then why does sizeof(a ? true : false) give an output of four bytes?


Here, ternary operator return boolean type,

OK, there's more to that!

In C, the result of this ternary operation is of type int . [notes below (1,2)]

Hence the result is the same as the expression sizeof(int) , on your platform.

Note 1: Quoting C11 , chapter §7.18, Boolean type and values <stdbool.h>

[....] The remaining three macros are suitable for use in #if preprocessing directives. They are

true

which expands to the integer constant 1,

false

which expands to the integer constant 0, [....]

Note 2: For conditional operator, chapter §6.5.15, ( emphasis mine )

The first operand is evaluated; there is a sequence point between its evaluation and the evaluation of the second or third operand (whichever is evaluated). The second operand is evaluated only if the first compares unequal to 0; the third operand is evaluated only if the first compares equal to 0; the result is the value of the second or third operand (whichever is evaluated), [...]

and

If both the second and third operands have arithmetic type, the result type that would be determined by the usual arithmetic conversions, were they applied to those two operands, is the type of the result. [....]

hence, the result will be of type integer and because of the value range, the constants are precisely of type int .

That said, a generic advice, int main() should better be int main (void) to be truly standard-conforming.


Here is a snippet from which is what included in the source

#ifndef __cplusplus

#define bool    _Bool
#define true    1
#define false   0

#else /* __cplusplus */

There macros true and false are declared as 1 and 0 respectively.

however in this case the type is the type of the literal constants. Both 0 and 1 are integer constants that fit in an int, so their type is int.

and the sizeof(int) in your case is 4.


Quick answer:

  • sizeof(a ? true : false) evaluates to 4 because true and false are defined in <stdbool.h> as 1 and 0 respectively, so the expression expands to sizeof(a ? 1 : 0) which is an integer expression with type int , that occupies 4 bytes on your platform. For the same reason, sizeof(true) would also evaluate to 4 on your system.

Note however that:

  • sizeof(a ? a : a) also evaluates to 4 because the ternary operator performs the integer promotions on its second and third operands if these are integer expressions. The same of course happens for sizeof(a ? true : false) and sizeof(a ? (bool)true : (bool)false) , but casting the whole expression as bool behaves as expected: sizeof((bool)(a ? true : false)) -> 1 .

  • also note that comparison operators evaluate to boolean values 1 or 0 , but have int type: sizeof(a == a) -> 4 .

The only operators that keep the boolean nature of a would be:

  • the comma operator: both sizeof(a, a) and sizeof(true, a) evaluate to 1 at compile time.

  • the assignment operators: both sizeof(a = a) and sizeof(a = true) have a value of 1 .

  • the increment operators: sizeof(a++) -> 1

Finally, all of the above applies to C only: C++ has different semantics regarding the bool type, boolean values true and false , comparison operators and the ternary operator: all of these sizeof() expressions evaluate to 1 in C++.


The ternary operator is a red herring.

    printf("%zu\n", sizeof(true));

prints 4 (or whatever sizeof(int) is on your platform).

The following assumes that bool is a synonym for char or a similar type of size 1, and int is larger than char .

The reason why sizeof(true) != sizeof(bool) and sizeof(true) == sizeof(int) is simply because true is not an expression of type bool . It's an expression of type int . It is #define d as 1 in stdbool.h .

There are no rvalues of type bool in C at all. Every such rvalue is immediately promoted to int , even when used as an argument to sizeof . Edit: this paragraph is not true, arguments to sizeof don't get promoted to int . This doesn't affect any of the conclusions though.


Regarding the boolean type in C

A boolean type was introduced fairly late in the C language, in the year 1999. Before then, C did not have a boolean type but instead used int for all boolean expressions. Therefore all logical operators such as > == ! etc return an int of value 1 or 0 .

It was custom for applications to use home-made types such as typedef enum { FALSE, TRUE } BOOL; , which also boils down to int -sized types.

C++ had a much better, and explicit boolean type, bool , which was no larger than 1 byte. While the boolean types or expressions in C would end up as 4 bytes in the worst case. Some manner of compatibility with C++ was introduced in C with the C99 standard. C then got a boolean type _Bool and also the header stdbool.h .

stdbool.h provides some compatibility with C++. This header defines the macro bool (same spelling as C++ keyword) that expands to _Bool , a type which is a small integer type, likely 1 byte large. Similarly, the header provides two macros true and false , same spelling as C++ keywords, but with backward compatibility to older C programs . Therefore true and false expand to 1 and 0 in C and their type is int . These macros are not actually of the boolean type like the corresponding C++ keywords would be.

Similarly, for backward compatibility purposes, logical operators in C still return an int to this day, even though C nowadays got a boolean type. While in C++, logical operators return a bool . Thus an expression such as sizeof(a == b) will give the size of an int in C, but the size of a bool in C++.

Regarding the conditional operator ?:

The conditional operator ?: is a weird operator with a couple of quirks. It is a common mistake to believe that it is 100% equivalent to if() { } else {} . Not quite.

There is a sequence point between the evaluation of the 1st and the 2nd or 3rd operand. The ?: operator is guaranteed to only evaluate either the 2nd or the 3rd operand, so it can't execute any side-effects of the operand that is not evaluated. Code like true? func1() : func2() will not execute func2() . So far, so good.

However , there is a special rule stating that the 2nd and 3rd operand must get implicitly type promoted and balanced against each other with the usual arithmetic conversions . ( Implicit type promotion rules in C explained here ). This means that the 2nd or 3rd operand will always be at least as large as an int .

So it doesn't matter that true and false happen to be of type int in C because the expression would always give at least the size of an int no matter.

Even if you would rewrite the expression to sizeof(a ? (bool)true : (bool)false) it would still return the size of an int !

This is because of implicit type promotion through the usual arithmetic conversions.





c11